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3 years ago

15

Following your push the ball rolls down the lane at 4.2m/s. What is the net force on the ball as it rolls down the lane at the c

onstant speed?

1 answer:

son4ous [18]3 years ago

3 0

Following your push the ball rolls down the lane at 4.2m/s. What is the net force on the ball as it rolls down the lane at the constant speed?

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KiRa [710]

Answer:

Probs paper cuz the rest are metals

4 0

1 year ago

If a person pulls back a rubber band on a slingshot without letting to go of it, what type of energy will the rubber band have?

Travka [436]

Answer:

Potential energy. Releasing it, the potential energy would convert into motion, kinetic energy.

Potential energy is when an object has some sort of potential eg. for motion such as in this example.

7 0

2 years ago

Consider the two vectors A = 3 î − ĵ and B = − î − 2 ĵ.

LUCKY_DIMON [66]

Answer:

(a) A+B = 2i-3j

(B) A-B = 4i + j

Explanation:

We have given two vectors A = 3i-j and B = -1-2j

We have to find the two vectors that is A+B and A-B

(A) In first art we have calculate A+B for this we have to add simply vector A and v ector B

So A+B = 3i-j-i-2j = 2i-3j

(B) In this part we have to find A-B for this we have to simply subtract B from A so A-B = 3i-j-(-i-2j) =3i-j+i+2j =4i+j

6 0

3 years ago

An object is thrown straight up with a velocity, in ft/s, given by v(t)= -32t + 83, where t is in seconds, from a height of 46 f

____ [38]

<h2>a) Initial velocity = 83 ft/s</h2><h2>b) Object's maximum speed = 99.4 ft/s</h2><h2>c) Object's maximum displacement = 153.64 ft</h2><h2>d) Maximum displacement occur at t = 2.59 seconds.</h2><h2>e) The displacement is zero when t = 5.70 seconds</h2><h2>f) Object's maximum height = 153.64 ft</h2>

Explanation:

We have velocity

v(t)= -32t + 83

Integrating

s(t) = -16t²+83t+C

At t = 0 displacement is 46 feet

46 = -16 x 0²+83 x 0+C

C = 46 feet

So displacement is

s(t) = -16t²+83t+46

a) Initial velocity is

v(0)= -32 x 0 + 83 = 83 ft/s

Initial velocity = 83 ft/s

b) Maximum velocity is when the object reaches ground, that is s(t) = 0 ft

Substituting

0 = -16t²+83t+46

t = 5.70 seconds

Substituting in velocity equation

v(t)= -32 x 5.70 + 83 = -99.4 ft/s

Object's maximum speed = 99.4 ft/s

c) Maximum displacement is when the velocity is zero

That is

-32t + 83 = 0

t = 2.59 s

Substituting in displacement equation

s(2.59) = -16 x 2.59²+83 x 2.59+46 = 153.64 ft

Object's maximum displacement = 153.64 ft

d) Maximum displacement occur at t = 2.59 seconds.

e) Refer part b

The displacement is zero when t = 5.70 seconds

f) Same as option d

Object's maximum height = 153.64 ft

5 0

3 years ago

One side of a triangle is increasing at a rate of 5 cm/sec and a second side is increasing at a rate of 7 cm/sec. If the area of

Vanyuwa [196]

Answer:

The rate of angle is 26.25 rad/sec.

Explanation:

Given that,

First side of triangle a= 20 cm

Second side of triangle b= 50 cm

One side of a triangle is increasing at a rate = 5 cm/sec

Second side is increasing at a rate = 7 cm/s

Angle $\theta=\dfrac{\pi}{3}=60^{\circ}$

If the area of the triangle remains constant,

We need to calculate rate of angle

Using formula of area of triangle

$A=\dfrac{1}{2}ab\sin\theta$

On differentiating

$\dfrac{dA}{dt}=\dfrac{1}{2}ab\cos\theta\dfrac{d\theta}{dt}+\dfrac{1}{2}a\sin\theta\dfrac{db}{dt}+\dfrac{1}{2}b\sin\theta\dfrac{da}{dt}$

Put the value into the formula

$0=\dfrac{1}{2}\times20\times50\cos60\dfrac{d\theta}{dt}+\dfrac{1}{2}\times20\sin60\times7+\dfrac{1}{2}\times50\sin60\times5$

$\dfrac{d\theta}{dt}=\dfrac{35\sqrt{3}\times\dfrac{25}{2}\sqrt{3}\times5}{250}$

$\dfrac{d\theta}{dt}=26.25\ rad/sec$

Hence, The rate of angle is 26.25 rad/sec.

8 0

2 years ago